1. ## An simple algebraic equation (x^2)

Hello everyone, I have got a quite nice equation to solve. I have worked some on it but it don't go pretty well.

The equation is:

$\frac { 1 - 3x^2 } { 4 - x^2 } = \frac { 9x - 1 } { 3x - 6 }$

My thoughts on how to solve it.

1 Get rid of the denominators

Multiply with both numerators.

2 Open up the parenthesises and put everything either left/right of '='.

That would result in $17x^2 - 23x = 0$

Going from here isn't too hard, but the anser I get is quite different from the books answer.

Thanks for all replies.

2. Originally Posted by λιεҗąиđ€ŗ
$\frac { 1 - 3x^2 } { 4 - x^2 } = \frac { 9x - 1 } { 3x - 6 }$

My thoughts on how to solve it.

1 Get rid of the denominators

Multiply with both numerators.

2 Open up the parenthesises and put everything either left/right of '='.

That would result in $17x^2 - 23x = 0$

Going from here isn't too hard, but the anser I get is quite different from the books answer.

Thanks for all replies.
Your method of attack sounds good.

So:
$\frac { 1 - 3x^2 } { 4 - x^2 } = \frac { 9x - 1 } { 3x - 6 }$

Let's multiply both sides by $(4 - x^2)(3x - 6)$:
$\frac { 1 - 3x^2 } { 4 - x^2 } \cdot (4 - x^2)(3x - 6) = \frac { 9x - 1 } { 3x - 6 } \cdot (4 - x^2)(3x - 6)$

$(1 - 3x^2 )(3x - 6) = (9x - 1)(4 - x^2)$

Let's take a moment here and do some factoring:
$3x - 6 = 3(x - 2)$
$4 - x^2 = (2 + x)(2 - x) = -(x + 2)(x - 2)$

Thus
$3(1 - 3x^2 )(x - 2) = -(9x - 1)(x + 2)(x - 2)$

Cancel the common x - 2:
$3(1 - 3x^2) = -(9x - 1)(x + 2)$

Now expand:
$3 - 9x^2 = -(9x^2 + 18x - x - 2);~x \neq -2$

$3 - 9x^2 = -9x^2 - 17x + 2;~x \neq -2$

$3 = -17x + 2;~x \neq -2$

$17x = -1;~x \neq -2$

$x = -\frac{1}{17};~x \neq -2$

Since $x \neq 2$ anyway, we can drop this part. Thus
$x = -\frac{1}{17}$

-Dan